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sammy [17]
4 years ago
14

13) What are the benefits and detriments of each of the following? Consider both the systems and the programmers’ levels. a. Sym

metric and asymmetric communication b. Automatic and explicit buffering c. Send by copy and send by reference d. Fixed-sized and variable-sized messages
Computers and Technology
1 answer:
dsp734 years ago
5 0

Answer:

Automatic and Explicit Buffering.

In the case of explicit buffering, the length of the queue is provided while in automatic buffering the queue size needs to be indefinite. In automatic buffering there is no need to block the sender while coping the message. While in explicit buffering the sender is blocked for the space in queue.

No memory is being wasted in explicit buffering.

Send by Copy and Send by Reference.

By using the send by copy method, the receiver is not able to change the state of parameter while send by reference allow. The advantage of using the send by reference method is that it allows to change a centralized application to its distributed version.

Fixed-sized and Variable-sized Messages.

In fixed size messaging refers, the buffer size is fixed. This means only a particular number of messages can only be supported by the fixed size buffer. The drawback of the fixed size messages is that they must be a part of fixed size buffer. These are not a part of variable size buffer. The advantage of variable size message is that the length of the message is variable means not fixed. The buffer length is unknown. The shared memory is being used by the variable size messages.

Explanation:

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See explaination

Explanation:

#input

o_t_n=int(input('Enter a string of octal digits: '))

#required variables

i = 1

dc = 0

#loop for conversion

while (o_t_n != 0):

#to find remainder

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# decimalToOctal.py

#input

d_c_n=int(input('Enter a decimal integer: '))

print("Quotient Remainder Octal")

#required variables

i = 1

o_c_n = 0

num=""

#loop for conversion

while (d_c_n != 0):

#to find remainder

rm = d_c_n % 8

d_c_n //= 8

o_c_n = o_c_n + rm * i

i *= 10

num = str(rm)+num

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